Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z^2 + 2z}{z^2 + 11z + 18} \times \dfrac{5z + 45}{z - 5} $
Answer: First factor the quadratic. $q = \dfrac{z^2 + 2z}{(z + 9)(z + 2)} \times \dfrac{5z + 45}{z - 5} $ Then factor out any other terms. $q = \dfrac{z(z + 2)}{(z + 9)(z + 2)} \times \dfrac{5(z + 9)}{z - 5} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ z(z + 2) \times 5(z + 9) } { (z + 9)(z + 2) \times (z - 5) } $ $q = \dfrac{ 5z(z + 2)(z + 9)}{ (z + 9)(z + 2)(z - 5)} $ Notice that $(z + 2)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 5z(z + 2)\cancel{(z + 9)}}{ \cancel{(z + 9)}(z + 2)(z - 5)} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $q = \dfrac{ 5z\cancel{(z + 2)}\cancel{(z + 9)}}{ \cancel{(z + 9)}\cancel{(z + 2)}(z - 5)} $ We are dividing by $z + 2$ , so $z + 2 \neq 0$ Therefore, $z \neq -2$ $q = \dfrac{5z}{z - 5} ; \space z \neq -9 ; \space z \neq -2 $